convert string to hex in python

Question

I have a script that calls a function that takes a hexadecimal number for an argument. The argument needs to the 0x prefix. The data source is a database table and is stored as a string, so it is returned '0x77'. I am looking for a way to take the string from the database and use it as an argument in hex form with the 0x prefix.

This works:

addr = 0x77
value = class.function(addr)

The database entry has to be a string, as most of the other records do not have hexadecimal values in this column, but the values could be changed to make it easier, so instead of '0x77', it could be '119'.

Answer 1

Your class.function expects an integer which can be represented either by a decimal or a hexadecimal literal, so that these two calls are completely equivalent:

class.function(0x77)
class.function(119)  # 0x77 == 119

Even print(0x77) will show 119 (because decimal is the default representation).

So, we should rather be talking about converting a string representation to integer. The string can be a hexadecimal representation, like '0x77', then parse it with the base parameter:

 >>> int('0x77', 16)
 119

or a decimal one, then parse it as int('119').

Still, storing integer whenever you deal with integers is better.

EDIT: as @gnibbler suggested, you can parse as int(x, 0), which handles both formats.

Answer 2

I think you're saying that you read a string from the database and you want to convert it to an integer, if the string has the 0x prefix you can convert it like so:

>>> print int("0x77", 16)
119

If it doesnt:

>>> print int("119")
119

Answer 3

>>> hex(119)
'0x77'
#or:
>>> hex(int("119"))
'0x77'

This should work for you.

You can also get the hex representation of characters:

>>> hex(ord("a"))
'0x61'